杭电oj 1006 Tick and Tick

Tick and Tick

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18985    Accepted Submission(s): 4871

Problem Description

The three hands of the clock are rotating every second and meeting each other many times everyday. Finally, they get bored of this and each of them would like to stay away from the other two. A hand is happy if it is at least D degrees from any of the rest. You are to calculate how much time in a day that all the hands are happy.

 

Input

The input contains many test cases. Each of them has a single line with a real number D between 0 and 120, inclusively. The input is terminated with a D of -1.

 

Output

For each D, print in a single line the percentage of time in a day that all of the hands are happy, accurate up to 3 decimal places.

 

Sample Input

012090-1

 

Sample Output

100.0000.0006.251

 

Author

PAN, Minghao

 

Source

ZJCPC2004

 

#include<iostream>#include<iomanip>#include<cstdio>//#define LOCALusing namespace std;double max(double a,double b,double c){    return a>b?(a>c?a:c):(b>c?b:c);}double min(double a,double b,double c){    return a<b?(a<c?a:c):(b<c?b:c);}int main(int argc,char **argv){    #ifdef LOCAL        freopen("data.in","r",stdin);        freopen("data.out","w",stdout);    #endif    int n;     //相差一度所要的时间    double sm = 10./59.;    double mh = 120./11.;    double sh = 120./719.;    //相差360度所要的时间    double _360sm = sm*360;    double _360mh = mh*360;    double _360sh = sh*360;    //快乐的总时间    double sum;    //快乐的开始时间和结束时间    double start_happy,end_happy;    //相差n度所要的时间,以及不再相差n度以上相差的时间    double _nsm,_nsh,_nmh,not_nsm,not_nsh,not_nmh;    while(cin>>n)    {        if(n==-1)            break;        sum = 0;        _nsm = n*sm;        _nsh = n*sh;        _nmh = n*mh;        not_nsm = _360sm-_nsm;        not_nsh = _360sh-_nsh;        not_nmh = _360mh-_nmh;        start_happy = max(_nsm,_nsh,_nmh);        end_happy = min(not_nsm,not_nsh,not_nmh);        while(start_happy<=43200&&end_happy<=43200)        {            start_happy = max(_nsm,_nsh,_nmh);            end_happy = min(not_nsm,not_nsh,not_nmh);            if(start_happy<end_happy)                sum += end_happy - start_happy;            if(end_happy == not_nsm)            {                _nsm += _360sm;                not_nsm += _360sm;            }else if(end_happy == not_nsh)            {                _nsh += _360sh;                not_nsh += _360sh;            }else{                _nmh += _360mh;                not_nmh += _360mh;            }        }        printf("%.3lf\n",sum/43200*100);    }    return 0;}